1 Answer
1

Note that $$\lim\sup f_n(x)=\inf_{n\geq 1} \sup_{k\geq n} f_k(x)$$
Let $g_n(x)=\sup_{k\geq n}f_k(x)$, then
$$g_n^{-1}(-\infty,a]=\lbrace x :g_n(x)\leq a\rbrace=\lbrace x :\sup_{k\geq n}f_k (x)\leq a\rbrace=\lbrace x: f_k(x)\leq a\mbox{ for all } k\geq n\rbrace$$
Hence $$g_n^{-1}(-\infty,a]=\bigcap_{k\geq n}f_k^{-1}(-\infty,a]$$
It follows that $g_n^{-1}(-\infty,a]$ is a measurable set (being intersection of measurable sets) and so $g_n$ is measurable..
In a similar fashion we can also prove that $\inf_{n\geq 1}g_n$ is also measurable (try), so we have $\lim\sup f_n(x)$ is measurable.

For the second part $f_n(x)=x^n$ converges pointwise to $g$ where $g(x)=0$ when $0\leq x<1$ and $g(1)=1$ and surely $g$ is not continous.